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This requires that the command:
expr "" \| ""
produce an empty line as its output.
In both BSD and System V historic practice, the | operator acted as a
boolean arithmetic operator like the &
operator. If the expressions evaluated to null strings, the result was "0\n",
not a null.
The rationale in clause E.4.22, P904, L6209-6213 acknowledges that
expr could be replaced by other
shell constructs in the POSIX.2 shell, but says the utility was kept
because of the many historical shell
scripts that use it. The rationale also mentions that other changes in early
drafts of the standard were
backed out because they weren't historic practice (see L6229-6230). It
seems strange that this change to
historical practice was not documented if it was intentional.
To match historic BSD and System V implementations, the description of
the | operator on P277, L4159-
4169 should have been something like:
expr1 | expr2 Returns the evaluation of expr1 if it is neither null nor zero;
otherwise, the evaluation ofexpr2
if it is not null; otherwise, returns zero.
Was this change to historic practice intentional?

Interpretation Response
The standard states the behavior for the | operator for the expr cmd, and
conforming implementations
must conform to this. However, concerns have been raised about this
which are being referred to the
sponsor.